Question: The side of the base of a square pyramid is increasing at a rate of $6$ meters per minute and the height of the pyramid is decreasing at a rate of $1$ meter per minute. At a certain instant, the base's side is $3$ meters and the height is $9$ meters. What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $-111$ (Choice B) B $105$ (Choice C) C $-105$ (Choice D) D $111$ The volume of a square pyramid with base side $s$ and height $h$ is $\dfrac{1}{3}s^2h$.
Explanation: Setting up the math Let... $s(t)$ denote the base side of the pyramid at time $t$, $h(t)$ denote the height of the pyramid at time $t$, and $V(t)$ denote the volume of the pyramid at time $t$. We are given that $s'(t)=6$ and $h'(t)=-1$ (notice that $h'$ is negative). We are also given that that $s(t_0)=3$ and $h(t_0)=9$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures The measures relate to each other through the formula for the volume of a square pyramid: $V(t)=\dfrac{1}{3}[s(t)]^2h(t)$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=\dfrac{2}{3}s(t)s'(t)h(t)+\dfrac{1}{3}[s(t)]^2h'(t)$ Using the information to solve Let's plug ${s(t_0)}={3}$, ${s'(t_0)}={6}$, ${h(t_0)}={9}$, and $C{h'(t_0)}=C{-1}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=\dfrac{2}{3}{s(t_0)}{s'(t_0)}{h(t_0)}+\dfrac{1}{3}[{s(t_0)}]^2C{h'(t_0)} \\\\ &=\dfrac{2}{3}({3})({6})({9})+\dfrac{1}{3}({3})^2(C{-1}) \\\\ &=105 \end{aligned}$ In conclusion, the rate of change of the volume of the pyramid at that instant is $105$ cubic meters per minute. Since the rate of change is positive, we know that the volume is increasing.